A) \[0.44\text{ }g\]
B) \[0.488\text{ }L\]
C) \[0.2242\text{ }L\]
D) \[0.244\text{ }L\]
Correct Answer: B
Solution :
[b] 1 mol of a gas at standard conditions (i,e, SATP) \[=24.4\text{ }L\] \[\underset{(\text{It contains two active H-atoms)}\text{.}}{\mathop{HC\equiv C-\overset{OH}{\mathop{\overset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}}}\,\] 1 mol of compound \[(Mw=70g)=2mol\] of \[{{C}_{2}}{{H}_{6}}(g)\] \[70.0\text{ }g\] of compound \[=2\times 24.4L\] of \[{{C}_{2}}{{H}_{6}}\] \[0.7\text{ }g\] of compound \[=\frac{2\times 24.4\times 0.7}{70}\] \[=0.488L\]
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